Checking signs in Kirchhoff's loop rule equations for a circuit

[willydavidjr]

I have a diagram on my website and I also attached it to this message. My question is I am wondering if my equation for this loop rule is correct. I am trying to solve it but I am having a problem about the signs I am getting..

This is my equation for this loop:


$$I_2=I_1+I_3$$

$$6V-2I_3-2I_2=0$$

$$1V-2I_3-I_1=0$$

Do you think the above equations are correct?I used the equations but I get different signs that doesn't match the circuit on the diagram...

This is the site of the circuit: www.geocities.com/willydavidjr/loop.html

 

[berkeman]

Last equation is +I1. Look at the voltage *increase* through R1 in that direction due to I1.

 

[willydavidjr]

I see. Do you think berkeman the second equation should be 6V - 2I3 + 2I2=0? What do you think?

 

[willydavidjr]

I don't think that the third equation should be +I1. Because we will get negative answer for I3 and I1. Meaning they have opposite direction and it shouldn't be!

 

[Curious3141]

I see. Do you think berkeman the second equation should be 6V - 2I3 + 2I2=0? What do you think?

The second equation is fine as you wrote it (why do you think the sign is wrong ?). Just amend the third equation as berkeman suggested.

When taking a loop across a resistance, if going with the labelled flow of current, use the minus sign. If going against the labelled flow of current, use the plus sign.

 

[willydavidjr]

Berkeman what I get for this three equation just as I amended the 3rd equation are I1= 1 amp, I2= 1 amp and I3= 1 amp. Am I correct? I expected to get a negative value for I1 but I didn't get it. All have positive values. Do you think all flow of currents are correct?

 

[berkeman]

How can they all equal 1A when your first equation says I2=I1+I3 ?

When you get the answers, just go back and put them on the original drawing, and see if all the voltages and currents add up.

 

[willydavidjr]

No I mean 2A for I2. Sorry my fault.

 

[Curious3141]

No I mean 2A for I2. Sorry my fault.

I replied to your PM. The answer is correct.